Answer
The required ${{n}^{th}}$ degree polynomial function is $3{{x}^{4}}-{{x}^{3}}-9{{x}^{2}}+159x-52$.
Work Step by Step
The general linear equation is $f\left( x \right)={{a}_{n}}\left( x-{{c}_{1}} \right)\left( x-{{c}_{2}} \right)\left( x-{{c}_{3}} \right)\left( x-{{c}_{4}} \right)$ , the roots are ${{c}_{1}}=-4,{{c}_{2}}=\frac{1}{3},{{c}_{3}}=2+3i$ , and ${{c}_{4}}=2-3i$.
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( x-{{c}_{1}} \right)\left( x-{{c}_{2}} \right)\left( x-{{c}_{3}} \right)\left( x-{{c}_{4}} \right) \\
& ={{a}_{n}}\left( x+4 \right)\left( x-\frac{1}{3} \right)\left( x-2-3i \right)\left( x-2+3i \right) \\
& ={{a}_{n}}\left( {{x}^{4}}-\frac{{{x}^{3}}}{3}-3{{x}^{2}}+53x-\frac{52}{3} \right)
\end{align}$
Now, compute the value of ${{a}_{n}}$.
Use $f\left( 1 \right)=100$
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( {{x}^{4}}-\frac{{{x}^{3}}}{3}-3{{x}^{2}}+53x-\frac{52}{3} \right) \\
& 100={{a}_{n}}\left( {{\left( 1 \right)}^{4}}-\frac{{{\left( 1 \right)}^{3}}}{3}-3{{\left( 1 \right)}^{2}}+53\left( 1 \right)-\frac{52}{3} \right) \\
& 100={{a}_{n}}\left( 1-\frac{1}{3}-3+53-\frac{52}{3} \right) \\
& {{a}_{n}}=3
\end{align}$
Substitute the value of ${{a}_{n}}$ in the linear equation.
$\begin{align}
& f\left( x \right)={{a}_{n}}\left( {{x}^{4}}-\frac{{{x}^{3}}}{3}-3{{x}^{2}}+53x-\frac{52}{3} \right) \\
& =3\left( {{x}^{4}}-\frac{{{x}^{3}}}{3}-3{{x}^{2}}+53x-\frac{52}{3} \right) \\
& =3{{x}^{4}}-{{x}^{3}}-9{{x}^{2}}+159x-52
\end{align}$
Therefore, the polynomial equation is $3{{x}^{4}}-{{x}^{3}}-9{{x}^{2}}+159x-52$.
